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Kotlin-based LeetCode algorithm problem solutions, regularly updated
2620\. Counter
Easy
Given an integer `n`, return a `counter` function. This `counter` function initially returns `n` and then returns 1 more than the previous value every subsequent time it is called (`n`, `n + 1`, `n + 2`, etc).
**Example 1:**
**Input:** n = 10 ["call","call","call"]
**Output:** [10,11,12]
**Explanation:**
counter() = 10 // The first time counter() is called, it returns n.
counter() = 11 // Returns 1 more than the previous time.
counter() = 12 // Returns 1 more than the previous time.
**Example 2:**
**Input:** n = -2 ["call","call","call","call","call"]
**Output:** [-2,-1,0,1,2]
**Explanation:** counter() initially returns -2. Then increases after each sebsequent call.
**Constraints:**
* `-1000<= n <= 1000`
* `At most 1000 calls to counter() will be made`